Solve the following differential equations: dy dx = x(2 +1) + y

Question

Solve the following differential equations:$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}(2\log\text{x}+1)}{\sin\text{y + y}\cos\text{y}}$

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Answer

we have, $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}(2\log\text{x}+1)}{\sin\text{y + y}\cos\text{y}}$
$\Rightarrow(\sin\text{y+y}\cos\text{y})\text{dy = x}(2\log\text{x}+1)\text{dx}$
Integrating both sides, we get
$\int(\sin\text{y+y}\cos\text{y})\text{dy}=\int\text{x}(2\log\text{x}+1)\text{dx}$
$\Rightarrow\int\sin\text{y dy}+\int\text{y}\cos\text{y dy }=2\int\text{x}\log\text{x dx}+\int\text{x dx}$
$\Rightarrow-\cos\text{y}+\Big[\text{y}\int\cos\text{y dy}-\int\Big\{\frac{\text{d}}{\text{dy}}\text{(y)}\int\cos\text{y dy}\Big\}\text{dy}\Big]\\=2\Big[\log\text{x}\int\text{x dx}-\int\Big\{\frac{\text{d}}{\text{dx}}(\log\text{x})\int\text{x dx}\Big\}\text{dx}\Big]+\frac{\text{x}^2}{2}$
$\Rightarrow-\cos\text{y}+\Big[\text{y}\sin\text{y}-\int\sin\text{y dy}\Big]=2\Big[\log\text{x}\times\frac{\text{x}^2}{2}-\int\frac{1}{\text{x}}\times\frac{\text{x}^2}{2}\Big]+\frac{\text{x}^2}{2}$
$\Rightarrow-\cos\text{y+y}\sin\text{y}+\cos\text{y}=\text{x}^2\log\text{x}-\frac{\text{x}^2}{2}+\frac{\text{x}^2}{2}+\text{C}$
$\Rightarrow\text{y}\sin\text{y}=\text{x}^2\log\text{x + C}$
Hence, $\text{y}\sin\text{y = x}^2\log\text{x + C}$ is the required solution.

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